Quantum Qual Problems: Difference between revisions
No edit summary |
No edit summary |
||
| Line 44: | Line 44: | ||
The eigenenergies <br> | The eigenenergies <br> | ||
<math>E_{n_x} = \frac{\pi^2 h^2 {n_x}^2}{2ma^2}</math> | <math>E_{n_x} = \frac{\pi^2 h^2 {n_x}^2}{2ma^2}</math>, <math>E_{n_y} = \frac{\pi^2 h^2 {n_y}^2}{2mb^2}</math>, <math>E_{n_z} = \frac{\pi^2 h^2 {n_z}^2}{2mc^2}</math> <br> | ||
Revision as of 03:37, 16 August 2007
1.) Given a quantum mechanical particle of mass [math]\displaystyle{ M }[/math] confined inside a box of sides [math]\displaystyle{ a,b,c }[/math]. The particle is allowed to move freely between [math]\displaystyle{ 0 <x < a, 0< y< b }[/math] and [math]\displaystyle{ 0<z<c }[/math].
- Use the time-independent Schrodinger equation for this problem to obtain the general form for the eigenfunctions of the particle
- Now apply boundary conditions to obtain the specific eigenfunctions and eigenenergies for this specific problem.
- Assume [math]\displaystyle{ a=b=c }[/math] and find the first 6 eigenenergies of the problem in terms of the box side length ([math]\displaystyle{ a }[/math]), the particle mass ([math]\displaystyle{ M }[/math]) and standard constants. What are their quantum number? Make a sketch of the eigenvalue spectrum, a table listing these eigenenergies and the quantum numbers of all the states that correspond to them.
Solution:
2.)
- [math]\displaystyle{ [- \frac{h^2}{2m}\Delta^2 + V]W(x,y,z)=E W(x,y,z) }[/math]
In our case, using separation of variables, we will get 3 differential equations for x, y and z. W(x,y,z)=w(x)w(y)w(z)
[math]\displaystyle{ \frac{d^2 w(x)}{dx^2} - m^2 w(x) = 0 }[/math](1)
The same will be for y and z.
Solution of equation (1) is following
[math]\displaystyle{ w(x) = A\sin(mx)+B\cos(mx) }[/math]
[math]\displaystyle{ w(y) = C\sin(ky)+D\cos(ky) }[/math]
[math]\displaystyle{ w(z) = E\sin(qz)+F\cos(qz) }[/math]
- Applying B. C. at x=y=z=0 wave function should be zero, that means B=D=F=0. We have
[math]\displaystyle{ w(x) = A\sin(mx) }[/math]
[math]\displaystyle{ w(y) = C\sin(ky) }[/math]
[math]\displaystyle{ w(z) = E\sin(qz) }[/math]
Also, w(a)=0 which gives [math]\displaystyle{ A\sin(ma)=0, m=\frac{\pi n_x}{a} }[/math]. For y component [math]\displaystyle{ C\sin(kb)=0, k=\frac{\pi n_y}{b} }[/math] and for z [math]\displaystyle{ E\sin(qc)=0, q=\frac{\pi n_z}{c} }[/math]
A, C and E are normalization constants
[math]\displaystyle{ \frac{1}{A^2}=\int\sin^2 (\pi nx/a) dx = \frac{a}{2} }[/math], limits are from 0 to a.
The eigenfunction for each component will be
[math]\displaystyle{ w(x) = \sqrt{\frac{2}{a}} \sin(\pi n_x x/a) }[/math]
[math]\displaystyle{ w(x) = \sqrt{\frac{2}{b}} \sin(\pi n_y y/b) }[/math]
[math]\displaystyle{ w(x) = \sqrt{\frac{2}{c}} \sin(\pi n_z z/c) }[/math]
The eigenenergies
[math]\displaystyle{ E_{n_x} = \frac{\pi^2 h^2 {n_x}^2}{2ma^2} }[/math], [math]\displaystyle{ E_{n_y} = \frac{\pi^2 h^2 {n_y}^2}{2mb^2} }[/math], [math]\displaystyle{ E_{n_z} = \frac{\pi^2 h^2 {n_z}^2}{2mc^2} }[/math]